JEE Main 2024 Chemistry question
The mass of sodium acetate (CH₃COONa) required to prepare 250 mL of 0.35 M aqueous solution is g. (Molar mass of CH₃COONa is 82.02 g mol⁻¹)
Answer: 7
Solution
Moles = 0.35 × 250/1000 = 0.0875 mol. Mass = 0.0875 × 82.02 = 7.18 g ≈ 7 g.